Problem: Simplify and expand the following expression: $ \dfrac{t - 7}{t + 8}-\dfrac{t}{5t + 1} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(t + 8)(5t + 1)$ Multiply the first term by $\dfrac{5t + 1}{5t + 1}$ $ \begin{align*} \dfrac{t - 7}{t + 8} \times \dfrac{5t + 1}{5t + 1} & = \dfrac{(t - 7)(5t + 1)}{(t + 8)(5t + 1)} \\ & = \dfrac{5t^2 - 34t - 7}{(t + 8)(5t + 1)}\end{align*} $ Multiply the second term by $\dfrac{t + 8}{t + 8}$ $ \begin{align*} \dfrac{t}{5t + 1} \times \dfrac{t + 8}{t + 8} & = \dfrac{(t)(t + 8)}{(5t + 1)(t + 8)} \\ & = \dfrac{t^2 + 8t}{(5t + 1)(t + 8)}\end{align*} $ Now we have: $ = \dfrac{5t^2 - 34t - 7}{(t + 8)(5t + 1)} - \dfrac{t^2 + 8t}{(5t + 1)(t + 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{5t^2 - 34t - 7 - (t^2 + 8t)}{(t + 8)(5t + 1)} $ $ = \dfrac{5t^2 - 34t - 7 - t^2 - 8t}{(t + 8)(5t + 1)} $ $ = \dfrac{4t^2 - 42t - 7}{(t + 8)(5t + 1)}$ Expand the denominator: $ = \dfrac{4t^2 - 42t - 7}{5t^2 + 41t + 8}$